Answer: $f(x) = x + 1$ and $g(x) = 1/x$.
The pair above is a solution. In order to prove that there is no other solution, fix $x \in \mathbb{R}_{>0}$ and denote $g^0 = x$ and $g^n = g(g^{n-1})$ for $n \ge 1$.
We have $g(x)(1 + g(g(x))) = f(g(x)) = f(x)g(x) = x(1 + g(x))g(x)$. Since $g(x) \ne 0$, we have $1 + g^2 = x(1 + g)$. This can be rewritten as
$$gg^2 - 1 = (xg - 1)(1 + g).$$
Then $g^2g^3 - 1 = (gg^2 - 1)(1 + g^2) = (xg - 1)(1 + g)(1 + g^2)$ and more generally
$$g^n g^{n+1} - 1 = (g^{n-1}g^n - 1)(1 + g^n) = (xg - 1)(1 + g)(1 + g^2) \dots (1 + g^n)$$
for any $n \ge 1$ by induction. There is $n > m$ with $g^n = g^m$ by assumption. Then we have
$$(xg - 1)(1 + g) \dots (1 + g^m) ((1 + g^{m+1}) \dots (1 + g^n) - 1) = 0.$$
Since $g^k > 0$, we must have $xg - 1 = 0$. It follows that $g = 1/x$ and $f = x + 1$.